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Question Number 49559 by Rio Michael last updated on 07/Dec/18

Commented by Rio Michael last updated on 07/Dec/18

The figure below shows an object of mass 60kg pulled up  by an effort of 4000J through an inclined plane 30° to the   horizontal.  a) Sketch the diagram showing all forces acting on the object  as it moves up the inclined plane.  b) The object is allowed to fall through a vertical height of 6m  calculate the change in potential energy.  c) Calculate the efficiency of the system above hence state  why its less then 100%.

$${The}\:{figure}\:{below}\:{shows}\:{an}\:{object}\:{of}\:{mass}\:\mathrm{60}{kg}\:{pulled}\:{up} \\ $$$${by}\:{an}\:{effort}\:{of}\:\mathrm{4000}{J}\:{through}\:{an}\:{inclined}\:{plane}\:\mathrm{30}°\:{to}\:{the}\: \\ $$$${horizontal}. \\ $$$$\left.{a}\right)\:{Sketch}\:{the}\:{diagram}\:{showing}\:{all}\:{forces}\:{acting}\:{on}\:{the}\:{object} \\ $$$${as}\:{it}\:{moves}\:{up}\:{the}\:{inclined}\:{plane}. \\ $$$$\left.{b}\right)\:{The}\:{object}\:{is}\:{allowed}\:{to}\:{fall}\:{through}\:{a}\:{vertical}\:{height}\:{of}\:\mathrm{6}{m} \\ $$$${calculate}\:{the}\:{change}\:{in}\:{potential}\:{energy}. \\ $$$$\left.{c}\right)\:{Calculate}\:{the}\:{efficiency}\:{of}\:{the}\:{system}\:{above}\:{hence}\:{state} \\ $$$${why}\:{its}\:{less}\:{then}\:\mathrm{100\%}. \\ $$

Answered by afachri last updated on 07/Dec/18

Commented by afachri last updated on 08/Dec/18

sir, pardon me. i did a mistake in the   diagram. the arrow of w cos30   should be w , and w arrow  should be   w cos30. i am so sorry.

$$\mathrm{sir},\:\mathrm{pardon}\:\mathrm{me}.\:\mathrm{i}\:\mathrm{did}\:\mathrm{a}\:\mathrm{mistake}\:\mathrm{in}\:\mathrm{the}\: \\ $$$$\mathrm{diagram}.\:\mathrm{the}\:\mathrm{arrow}\:\mathrm{of}\:{w}\:{cos}\mathrm{30}\: \\ $$$$\mathrm{should}\:\mathrm{be}\:{w}\:,\:\mathrm{and}\:{w}\:\mathrm{arrow}\:\:\mathrm{should}\:\mathrm{be}\: \\ $$$${w}\:{cos}\mathrm{30}.\:\mathrm{i}\:\mathrm{am}\:\mathrm{so}\:\mathrm{sorry}. \\ $$$$ \\ $$

Commented by Rio Michael last updated on 08/Dec/18

no problem sir

$${no}\:{problem}\:{sir} \\ $$

Commented by Rio Michael last updated on 08/Dec/18

thanks sir i saw that already

$${thanks}\:{sir}\:{i}\:{saw}\:{that}\:{already} \\ $$

Commented by afachri last updated on 07/Dec/18

your welcome Sir. the other  answer below.

$$\mathrm{your}\:\mathrm{welcome}\:\mathrm{Sir}.\:\mathrm{the}\:\mathrm{other} \\ $$$$\mathrm{answer}\:\mathrm{below}. \\ $$

Answered by afachri last updated on 07/Dec/18

B) let′s  say the gravity conts : 10 m/s^2         △Potential energy  =  m.g.△h                                                            =  (60 Kg)(10  m/s^2 )(6 m)                                                            =  3600 J  C)  find first :   w sin(30) and F                      w sin(30°)  =  m × g × sin(30°)                                                =  60  ×  10  (1/(2 ))                                                =  300 J          and,                           F                  =    (W/s)                                                   = (4000 J) / 12 m = ((1000)/3) N         so, the efficiency of system will be :        e  =  ((w sin (30°))/F) × 100%             =  90%  ⇒  reason : cause applied effort (4000 J ; 300 N)                                       is over than the force needed (((1000)/3) N)

$$\left.\boldsymbol{\mathrm{B}}\right)\:\mathrm{let}'\mathrm{s}\:\:\mathrm{say}\:\mathrm{the}\:\mathrm{gravity}\:\mathrm{conts}\::\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\bigtriangleup\boldsymbol{{Potential}}\:\boldsymbol{{energy}}\:\:=\:\:\boldsymbol{{m}}.\boldsymbol{{g}}.\bigtriangleup\boldsymbol{{h}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\left(\mathrm{60}\:\mathrm{Kg}\right)\left(\mathrm{10}\:\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \right)\left(\mathrm{6}\:\mathrm{m}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{3600}\:\mathrm{J} \\ $$$$\left.\boldsymbol{\mathrm{C}}\right)\:\:\mathrm{find}\:\mathrm{first}\::\:\:\:{w}\:{sin}\left(\mathrm{30}\right)\:\mathrm{and}\:{F} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{w}\:{sin}\left(\mathrm{30}°\right)\:\:=\:\:{m}\:×\:{g}\:×\:{sin}\left(\mathrm{30}°\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{60}\:\:×\:\:\mathrm{10}\:\:\frac{\mathrm{1}}{\mathrm{2}\:} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{300}\:\mathrm{J} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{and}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{F}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\:\:\frac{{W}}{{s}}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\mathrm{4000}\:\boldsymbol{\mathrm{J}}\right)\:/\:\mathrm{12}\:\mathrm{m}\:=\:\frac{\mathrm{1000}}{\mathrm{3}}\:\mathrm{N} \\ $$$$\:\:\:\:\:\:\:\mathrm{so},\:\mathrm{the}\:\mathrm{efficiency}\:\mathrm{of}\:\mathrm{system}\:\mathrm{will}\:\mathrm{be}\:: \\ $$$$\:\:\:\:\:\:{e}\:\:=\:\:\frac{{w}\:{sin}\:\left(\mathrm{30}°\right)}{{F}}\:×\:\mathrm{100\%} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{90\%}\:\:\Rightarrow\:\:\mathrm{reason}\::\:\mathrm{cause}\:\mathrm{applied}\:\mathrm{effort}\:\left(\mathrm{4000}\:\mathrm{J}\:;\:\mathrm{300}\:\mathrm{N}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{is}\:\mathrm{over}\:\mathrm{than}\:\mathrm{the}\:\mathrm{force}\:\mathrm{needed}\:\left(\frac{\mathrm{1000}}{\mathrm{3}}\:\mathrm{N}\right) \\ $$$$ \\ $$$$ \\ $$

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